# One liter of solution was prepared by dissolving 0.025 mol of formic acid, HCHO 2 , and 0.018 mol of sodium formate, NaCHO 2 , in water. What was the pH of the solution? K a for formic acid is 1.7 × 10 −4 .

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

#### Solutions

Chapter
Section
Chapter 16.5, Problem 16.11E
Textbook Problem

## One liter of solution was prepared by dissolving 0.025 mol of formic acid, HCHO2, and 0.018 mol of sodium formate, NaCHO2, in water. What was the pH of the solution? Ka for formic acid is 1.7 × 10−4.

Expert Solution
Interpretation Introduction

Interpretation:

The pH of the solution prepared by dissolving 0.025 mol of formic acid and 0.018 mol of sodium formate in water has to be calculated

Concept Information:

The equilibrium expression for the weak acid is given below.

Ka=[H+][A-][HA]

Where,

Ka is acid ionization constant,

[H+] is concentration of hydrogen ion

[A-] is concentration of acid anion

[HA] is concentration of the acid

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion [H+] concentration.

pH=-log[H+]

To Calculate: The pH of the solution prepared by dissolving 0.025 mol of formic acid and 0.018 mol of sodium formate in water

### Explanation of Solution

Given data:

The amount of formic acid = 0.025 mol

The amount of sodium formate = 0.018 mol

The Ka for formic acid is 1.7×104

Identify the reaction for acid and find out whether any ions in solution from added salt sodium formate that will disturb the chemical reaction

HCOOH + H2O HCOO- + H+

Sodium formate gives Na+ and formate ion (HCOO-)

Assemble the usual table using a starting formate ion (CHO2-) of 0.018 mol NaCHO2

Construct the equilibrium table, by considering the concentration of formate ion from sodium formate

 HCOOH   +   H2O  ⇌   HCOO-   +   H+ Initial (M) 0.025 −x 0.025-x 0.018 0.00 Change (M) +x +x Equilibrium (M) 0

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